Question

What happens when:
(i) ${\text{S}}{{\text{O}}_{\text{2}}}$ gas is passed through an aqueous solution of ${\text{F}}{{\text{e}}^{3 + }}$ salt?
(ii) ${\text{Xe}}{{\text{F}}_{\text{4}}}$ reacts with ${\text{Sb}}{{\text{F}}_{\text{5}}}$?

Answer 1

Sulphur dioxide gas is a reducing agent thus, it reduces the species it comes in contact with and decreases its oxidation state. During the reaction of ${\text{Xe}}{{\text{F}}_{\text{4}}}$ with ${\text{Sb}}{{\text{F}}_{\text{5}}}$, migration of one fluorine ion takes place.

Complete step by step answer:
(i) When ${\text{S}}{{\text{O}}_{\text{2}}}$ gas is passed through an aqueous solution of ${\text{F}}{{\text{e}}^{3 + }}$ salt, the ${\text{F}}{{\text{e}}^{3 + }}$ ions are converted to ${\text{F}}{{\text{e}}^{2 + }}$ ions.
The oxidation state changes form 3+ to 2+. Thus, the oxidation decreases by 1.
A decrease in oxidation state is known as reduction. Thus, the ${\text{F}}{{\text{e}}^{3 + }}$ ions reduce to ${\text{F}}{{\text{e}}^{2 + }}$ ions when ${\text{S}}{{\text{O}}_{\text{2}}}$ gas is passed through its solution.
The reaction when ${\text{S}}{{\text{O}}_{\text{2}}}$ gas is passed through an aqueous solution of ${\text{F}}{{\text{e}}^{3 + }}$ salt is as follows:
${\text{2F}}{{\text{e}}^{3 + }} + {\text{S}}{{\text{O}}_{\text{2}}} + 2{{\text{H}}_2}{\text{O}} \to {\text{2F}}{{\text{e}}^{2 + }} + {\text{SO}}_4^{2 - } + 4{{\text{H}}^ + }$
Thus, when ${\text{S}}{{\text{O}}_{\text{2}}}$ gas is passed through an aqueous solution of ${\text{F}}{{\text{e}}^{3 + }}$ salt, the ${\text{F}}{{\text{e}}^{3 + }}$ ions are reduced to ${\text{F}}{{\text{e}}^{2 + }}$ ions.
(ii) When ${\text{Xe}}{{\text{F}}_{\text{4}}}$ reacts with ${\text{Sb}}{{\text{F}}_{\text{5}}}$ one fluorine atom from ${\text{Xe}}{{\text{F}}_{\text{4}}}$ migrates to ${\text{Sb}}{{\text{F}}_{\text{5}}}$. Thus, ${\text{Xe}}{{\text{F}}_{\text{4}}}$ is converted to ${\text{Xe}}{{\text{F}}_{\text{3}}}$ and ${\text{Sb}}{{\text{F}}_{\text{5}}}$ is converted to ${\text{Sb}}{{\text{F}}_{\text{6}}}$.
The reaction is as follows:
${\text{Xe}}{{\text{F}}_{\text{4}}} + {\text{Sb}}{{\text{F}}_{\text{5}}} \to {\text{Xe}}{{\text{F}}_{\text{3}}} + {\text{Sb}}{{\text{F}}_{\text{6}}}$
Thus, when ${\text{Xe}}{{\text{F}}_{\text{4}}}$ reacts with ${\text{Sb}}{{\text{F}}_{\text{5}}}$, ${\text{Xe}}{{\text{F}}_{\text{3}}}$ and ${\text{Sb}}{{\text{F}}_{\text{6}}}$ are formed.

Note: Sulphur dioxide is a toxic gas. The main source of sulphur dioxide is volcanic activity and the combustion of fossil fuels that are contaminated by sulphur compounds. It is used in production of sulphuric acid. It is used as a preservative for preserving dried fruits. It is also used in winemaking. It is an extremely good reducing agent.
${\text{Xe}}{{\text{F}}_{\text{4}}}$ is known as xenon tetrafluoride. The geometry of xenon fluoride is octahedral because it has two lone pairs around the xenon atom.